Question

Let X1, ... , Xn be independent uniform (0, 1) random variables. Let R = X(n) − X(1) denote the range and M = [X(n) + X(1)]/2 the midrange of X1, ..., Xn. Compute the joint density function of R and M.

Short answer

$f(R,M)=\left\{\begin{array}{l}1;0\le (r,m)\le 1\\ 0;\mathrm{otherwise}\end{array}\right.$

Step-by-step solution

Introduction

Let $X_1,\dots ,X_n$ be independent uniform random variables.
The range of the uniform distribution is defined as $R=X_{\left(n\right)}-X_{\left(1\right)}$

The Midrange of the uniform distribution is defined as

$M=\frac{\left[X_{\left(n\right)}+X_{\left(1\right)}\right]}{2}$

Explanation

Compute the joint density function of R and M
The probability density function of the Uniform distribution is given by,

$$\begin{gathered} f_i\left(x_i\right)=\left\{\begin{array}{l}1;0\le x_i\le 1\\ 0;Otherwise\end{array}\right. \\ Sinceallthevariablesareindependenttoeachother, \\ thejointdistributionisgivenby, \\ f\left(x_1{x}_n\right)=f\left(x_1\right)f\left(x_n\right)=1 \\ Consider \\ R=X_{\left(n\right)}-X_{\left(1\right)}M=\frac{X_{\left(n\right)}+X_{\left(1\right)}}{2} \end{gathered}$$

Explanation

Let use say x, y are the minimum and maximum values in a sample which is taken from the given distribution. That is $X_{\left(n\right)}=y$and $X_{\left(1\right)}=x$
By using the transformation,

$$\begin{gathered} r=y-x\Rightarrow x=y-r \\ m=\frac{y+x}{2} \\ \Rightarrow y=2m-x \\ \Rightarrow y=2m-(y-r) \\ \Rightarrow 2y=2m+r \\ \Rightarrow y=m+\frac{r}{2} \\ And, \\ x\&=y-r \\ =m+\frac{r}{2}-r \\ =m-\frac{r}{2} \\ Thus, \\ x=m-\frac{r}{2}y=m+\frac{r}{2} \end{gathered}$$

Explanation

Use Jacobian transformation to find the Jacobian coefficient. That is,

$$\begin{gathered} J=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial m}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial m}\end{array}\right| \\ =\left|\begin{array}{cc}\frac{\partial }{\partial r}\left(m-\frac{r}{2}\right)& \frac{\partial }{\partial m}\left(m-\frac{r}{2}\right)\\ \frac{\partial }{\partial r}\left(m+\frac{r}{2}\right)& \frac{\partial }{\partial m}\left(m+\frac{r}{2}\right)\end{array}\right| \\ =\left|\begin{array}{cc}-\frac{1}{2}& 1\\ \frac{1}{2}& 1\end{array}\right| \\ =-\frac{1}{2}\left(1\right)-\frac{1}{2}\left(1\right) \\ =-1 \end{gathered}$$

Explanation 

$$\begin{gathered} Now,thejointdistributionofRandMisgivenby, \\ f(R,M)=f(x_1,x_n)\times \frac{1}{\left|J\right|} \\ =1\times \frac{1}{|-1|} \\ =1 \\ Therefore,thejointdensityfunctionofRandMis, \\ f(R,M)=\left\{\begin{array}{l}1;0\le (r,m)\le 1\\ 0;\mathrm{otherwise}\end{array}\right. \end{gathered}$$